City Subway Survey and Rare-Event Approximations: 04-BS-2, December 2019, Q2

Question 2: City Subway Survey and Rare-Event Approximations (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. 2(A): p = 0.70 of adults favour a third subway line. 2(B): p = 0.0012 probability of a serious accident, n = 2,500.

Find. 2(A)(a) P(4<X<8), n=15; (b) P(fewer than 3 not in favour), n=10; (c) P(X<2,750), n=4,000 via normal approximation. 2(B) P(X<3) via Poisson approximation, with justification.

Approach. (A)(a)-(b) use the exact Binomial pmf/cdf (n small); (A)(c) uses the normal approximation to the binomial (n large, np and nq both >5) with a continuity correction; (B) uses the Poisson approximation to the binomial (n large, p small, np moderate).

2(A)(a) n=15, "more than four but fewer than eight" = P(X∈{5,6,7}). With $X\sim\text{Binomial}(15,0.70)$, $$P(5\le X\le 7)=\sum_{k=5}^{7}\binom{15}{k}(0.70)^k(0.30)^{15-k}=0.00298+0.01159+0.03477$$ $$\boxed{P(4\lt X\lt 8)=0.0493}$$
2(A)(b) n=10, "fewer than three not in favour." Let Y = number NOT in favour, so $Y\sim\text{Binomial}(10,0.30)$ (q = 1−0.70 = 0.30). $$P(Y\lt 3)=P(Y\le 2)=\sum_{k=0}^{2}\binom{10}{k}(0.30)^k(0.70)^{10-k}$$ $$\boxed{P(Y\lt 3)=0.3828}$$
2(A)(c) n=4,000, normal approximation, P(X<2,750). Since np=2,800>5 and nq=1,200>5, $X\sim\text{Binomial}(4000,0.70)\approx N(np,\,npq)$: $$\mu=4000(0.70)=2{,}800, \qquad \sigma=\sqrt{4000(0.70)(0.30)}=28.98$$ With the continuity correction, $z=\dfrac{2749.5-2800}{28.98}=-1.74$, so $$\boxed{P(X\lt 2{,}750)\approx P(Z\lt -1.74)=0.0407}$$
2(B) n=2,500, p=0.0012, Poisson approximation, P(X<3). Since n is large and p is small with $\lambda=np=2500(0.0012)=3.0$ moderate, the Poisson approximation to the binomial applies (Binomial is well approximated by Poisson when $n\ge 20$ and $p\le 0.05$, here comfortably satisfied). $$P(X\lt 3)=P(0)+P(1)+P(2)=e^{-3}\left(1+3+\dfrac{3^2}{2!}\right)$$ $$\boxed{P(X\lt 3)=0.4232}$$ The normal approximation used in 2(A)(c) would be inappropriate here because np=3 is far below the np,nq>5 rule of thumb (nq=2497 is fine but np=3 is not) — the Poisson approximation is the correct tool for a rare event (small p) observed over many trials.

Question 2 — summary of results
Part	Quantity	Result
2(A)(a)	P(4<X<8), n=15	0.0493
2(A)(b)	P(fewer than 3 not in favour), n=10	0.3828
2(A)(c)	P(X<2,750), n=4,000 (normal approx.)	0.0407
2(B)	P(X<3), n=2,500 (Poisson approx., λ=3)	0.4232

Topic: Binomial distribution and its two classical large-sample approximations.

Key relations: normal approximation $X\approx N(np,npq)$ valid when np, nq > 5, always with a ±0.5 continuity correction; Poisson approximation $X\approx\text{Poisson}(np)$ valid when n is large and p is small (np moderate).

Why this works: the normal approximation follows from the CLT applied to a sum of Bernoulli trials; the Poisson approximation is the limiting case of the binomial as $n\to\infty$, $p\to 0$ with np held fixed (rare-event limit).

Common pitfall: applying the normal approximation to a small-p/large-n problem (as in 2(B), where np=3) gives an inaccurate answer because nq>5 alone is not sufficient — both np and nq must clear the threshold.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 5 (Discrete Distributions and their Approximations).

Question 2 of 8: City Subway Survey and Rare-Event Approximations

Question 2: City Subway Survey and Rare-Event Approximations (20 marks)